Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 11/30/24 4:03 AM, WM wrote:

On 29.11.2024 23:05, Richard Damon wrote:

On 11/29/24 4:39 PM, WM wrote:

On 29.11.2024 21:07, Richard Damon wrote:

On 11/29/24 2:28 PM, WM wrote:

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Analysis is no a lie.

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Bad Analysis, like what you do is.

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The limit of the infinite sequence 1/9, 1/9, 1/9, ... is 1/9. Nothing is clearer than that.
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Which doesn't actually mean anything.
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The limit x-> 0 of 0^x is 0 (as it is for all x)
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The limit x->0 of x^0 is 1 (as it is for all x)
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But 0^0 isn't defined, even though both of the limits seem to apprach it.

 1/9, 1/9, 1/9, ... has the unique limit 1/9.

So 1 == 0 by your logic and your world blows up.
0^x is 0, 0, 0, 0, ...  == 0 which then must be the value of 0^0
x^0 is 1, 1, 1, 1, ... == 1 which then must also be the value of 0^0
Thus, in your logic, 0 equal 1.

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You can only use the limit as the final result if it actually applies.
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Since the Infinite set is actually the same kind as any of the finite sets in the sequence, the limit doesn't apply

 It does.

Only of 0 == 1.
Sorry, but the properties of an infinite sequence can not be derived from the properties of finite sequences that just look sort of like it.
Your logic has gone BOOM and created all your darkness.

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Further>>>> I am using a very simple and sound rule. If all hats of finite

intervals (0, n] fail to cover more than 1/10, then it is impossible to cover more than 1/10 of the whole set ℕ because beyond all finite intervals and all finite n, there is no supply of black hats.

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So, you admit to MAKING UP your rules based on your own ideas

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This chain of arguing is irrefutable by consistent thinking.

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No, it is based on inconsistent thinking

 That is yours.

Nope, mine knows that 0 and 1 are different, and doesn't think they are the same.
I don't get to try to pick and choose which answers I will accept, but the answers are actually DEFINED, so real truth exists.

 Regards, WM
 

Sorry, you are just proving how stupid you are.